Question

The expression of the trajectory of a projectile is given as $y = px - qx^2$ , where $y$ and $x$ are respectively the vertical and horizontal displacements and $p$ and $q$ are constants. The time of flight of the projectile is

• A. $\frac{p^{2}}{4q}$
• B. $\frac{p^{2}}{2q}$
• C. $\sqrt{\frac{2p}{qg}}$
• D. $p\sqrt{\frac{2}{qg}}$

Answer 1

Answer: (D)

$p\sqrt{\frac{2}{qg}}$

Answer 2

Given, $y = px - qx^{2}$
$y_{max} when \frac{dy}{dx} = 0$
$\Rightarrow p - 2qx = 0$
$\Rightarrow y_{max}$ or max height $\left(H\right) = \frac{p^{2}}{4q}$
Now, $H = y_{max} = \frac{u^{2}_{y}}{2g}$
$\Rightarrow \frac{u_{y}^{2}}{2g} = \frac{p^{2}}{4q}$
$\Rightarrow u_{y} = \sqrt{\frac{gp^{2}}{2q}}$
Also, $T =$ time of flight
$= \frac{2u\, sin\, \theta}{g} = \frac{2u_{y}}{g}$
$= p\sqrt{\frac{2}{gq}}$